Sqlite选择值作为其他表的id

Question

I am working with SqLite, and I need connect table A with table B where the key is guid written as ["guid"] (TEXT). How do I accomplish that? (Should I use a substring value?)

Table A

|name| value    
|a   | ["13889483-d92a-483e-9e16-471cb22b82a3","a7ced9c5-e7bc-4214-be77-a26d8f86844b"]

Table B

| name                                 | value |
|13889483-d92a-483e-9e16-471cb22b82a3  | abc

Answer 1

You need to fix your table A so it is in First Normal Form :

Name  Value
a     13889483-d92a-483e-9e16-471cb22b82a3
a     a7ced9c5-e7bc-4214-be77-a26d8f86844b

Then you can connect the tables using a SQL join:

SELECT B.* FROM A
JOIN B ON B.Name = A.Value
WHERE A.Name = 'a'